Tính các tích phân sau:
a) \(\int_{0}^{1}(1+3x)^{\frac{3}{2}}dx\);
b) \(\int_{0}^{\frac{1}{2}}\frac{x^{3}-1}{x^{2}-1}dx\);
c) .
Câu a:
Đặt u = 1 + 3x ta có: du = 3dx
Khi x = 0 thì u = 1; khi x = 1 thì u = 4.
Do đó: \(\int_{0}^{1}(1+3x)^{\frac{3}{2}}dx =\frac{1}{3}\int_{0}^{4}u^{\frac{3}{2}}du\)
\(=\frac{1}{3}\frac{2}{5}u^{\frac{5}{2}} \Bigg|_{0}^{4}= \frac{2}{15}(4^{\frac{5}{2}}-1)=\frac{62}{15}\)
Câu b:
\(= \int_{0}^{\frac{1}{2}}\frac{x^2+x+1}{x+1}dx=\int_{0}^{\frac{1}{2}}(x+\frac{1}{x+1})dx\)
\(=\frac{x^{2}}{2}\Bigg|_{0}^{\frac{1}{2}}+ln\left | x+1 \right | \Bigg|_{0}^{\frac{1}{2}}=\frac{1}{8}+ln\frac{3}{2}\)
Câu c:
\(\left\{\begin{matrix} u=ln(1+x)\\ dv=\frac{dx}{x^2} \end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{1}{1+x}dx\\ v=-\frac{1}{x} \end{matrix}\right.\)
Ta có:
\(\int_{1}^{2}\frac{ln(1+x)}{x^{2}}dx =-\frac{1}{x}ln(1+x) \Bigg|^2_1+ \int_{1}^{2} \frac{dx}{x(1+x)}\)
\(=-\frac{1}{2}ln 3+ln 2 +\int_{1}^{2}\frac{dx}{x(1+x)}\)
Xét \(\int_{1}^{2}\frac{dx}{(x+1)x}\). Ta có: \(\frac{1}{x(1+x)}=\frac{1}{x}-\frac{1}{x+1}\)
Do đó:
\(\int_{1}^{2}\frac{dx}{x(x+1)}=\int_{1}^{2}\left ( \frac{1}{x}- \frac{1}{x+1} \right )dx=\int_{1}^{2}\frac{dx}{x}-\int_{1}^{2}\frac{dx}{x+1}\)
\(=lnx \Bigg|^2_1-ln(1+x)\Bigg|^2_1=ln2 -ln3+ln2\)
Vậy \(\int_{1}^{2}\frac{ln(1+x)}{x^2}dx=-\frac{1}{2}ln3 +ln2 +ln2-ln3+ln2\)
\(=3ln2-\frac{3}{2}ln3=ln8-\frac{3}{2}ln3=3ln \frac{2\sqrt{3}}{3}\)
-- Mod Toán 12
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